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circumcevian reflections

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  • Floor en Lyanne van Lamoen
    Dear friends, First of all I wish you a very happy and fruitful 2005. To start the new year let me offer a little theorem: If A B C is a circumcevian
    Message 1 of 8 , Jan 5, 2005
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      Dear friends,

      First of all I wish you a very happy and fruitful 2005.

      To start the new year let me offer a little theorem:

      If A'B'C' is a circumcevian triangle, and A"B"C" are the reflections of
      A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C), (A"BC")
      and (AB"C") are concurrent in one point.

      Kind regards,
      Floor.
    • Bernard Gibert
      Dear Floor and friends, may I wish all the best to all of you for this new year. ... your point is the isogonal conjugate of the infinite point of the
      Message 2 of 8 , Jan 5, 2005
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        Dear Floor and friends,

        may I wish all the best to all of you for this new year.

        > [FvL]To start the new year let me offer a little theorem:
        >
        > If A'B'C' is a circumcevian triangle [of point P], and A"B"C" are the
        > reflections of
        > A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
        > (A"BC")
        > and (AB"C") are concurrent in one point.

        your point is the isogonal conjugate of the infinite point of the
        direction which is perpendicular to HP.

        notice that ABC and A"B"C" are perspective iff P lies on the Jerabek
        hyperbola or at infinity.

        Best regards

        Bernard

        [Non-text portions of this message have been removed]
      • Floor en Lyanne van Lamoen
        Dear Antreas, ... No, you mixed up with (A BC), (AB C) and (ABC ) are concurrent in one point (H). Kind regards, Floor.
        Message 3 of 8 , Jan 5, 2005
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          Dear Antreas,

          > [FvL]:
          > >If A'B'C' is a circumcevian triangle, and A"B"C" are the reflections of
          > >A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C), (A"BC")
          > >and (AB"C") are concurrent in one point.
          >
          > Isn't it equivalent to this:
          >
          > The circumcircle of ABC and its reflections on the sidelines
          > of ABC are concurrent ?

          No, you mixed up with (A"BC), (AB"C) and (ABC") are concurrent in one point
          (H).

          Kind regards,
          Floor.
        • Floor en Lyanne van Lamoen
          Dear Bernard, ... [BG] ... Thanks, Bernard! The locus of the perspector for P on the Jerabek hyperbola is a quartic, the isogonal conjugate of the conic SUM
          Message 4 of 8 , Jan 6, 2005
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            Dear Bernard,

            > > [FvL]To start the new year let me offer a little theorem:
            > >
            > > If A'B'C' is a circumcevian triangle [of point P], and A"B"C" are the
            > > reflections of
            > > A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
            > > (A"BC")
            > > and (AB"C") are concurrent in one point.

            [BG]
            > your point is the isogonal conjugate of the infinite point of the
            > direction which is perpendicular to HP.
            >
            > notice that ABC and A"B"C" are perspective iff P lies on the Jerabek
            > hyperbola or at infinity.

            Thanks, Bernard!

            The locus of the perspector for P on the Jerabek hyperbola is a quartic, the
            isogonal conjugate of the conic

            SUM b^4c^4SA(b^2-c^2)^4x^2 + a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0
            cyclic

            Kind regards,
            Floor.
          • Bernard Gibert
            Dear Floor, ... it seems that your conic decomposes into two secant lines at X110, passing through X112 and X351. please do check this Best regards Bernard
            Message 5 of 8 , Jan 6, 2005
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              Dear Floor,

              > [FvL] The locus of the perspector for P on the Jerabek hyperbola is a
              > quartic, the
              > isogonal conjugate of the conic
              >
              > SUM b^4c^4SA(b^2-c^2)^4x^2 + a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0
              > cyclic

              it seems that your conic decomposes into two secant lines at X110,
              passing through X112 and X351.

              please do check this

              Best regards

              Bernard

              [Non-text portions of this message have been removed]
            • Floor en Lyanne van Lamoen
              Dear Bernard, ... [BG] ... This is caused by a typo. The conic should be SUM b^4c^4SA(b^2-c^2)^4x^2 - a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0 cyclic So with - in
              Message 6 of 8 , Jan 6, 2005
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                Dear Bernard,

                > > [FvL] The locus of the perspector for P on the Jerabek hyperbola is a
                > > quartic, the
                > > isogonal conjugate of the conic
                > >
                > > SUM b^4c^4SA(b^2-c^2)^4x^2 + a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0
                > > cyclic

                [BG]
                > it seems that your conic decomposes into two secant lines at X110,
                > passing through X112 and X351.
                >
                > please do check this

                This is caused by a typo. The conic should be

                SUM b^4c^4SA(b^2-c^2)^4x^2 - a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0
                cyclic

                So with - in stead of +. My apologies.

                Kind regards,
                Floor.
              • jpehrmfr
                Dear Floor ... Happy New Year to you and to each Hyacinthist. ... reflections of ... (A BC ) ... Note that if A ,B ,C lie on the circumcircle, the circles
                Message 7 of 8 , Jan 6, 2005
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                  Dear Floor
                  > First of all I wish you a very happy and fruitful 2005.

                  Happy New Year to you and to each Hyacinthist.

                  > To start the new year let me offer a little theorem:
                  > If A'B'C' is a circumcevian triangle, and A"B"C" are the
                  reflections of
                  > A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
                  (A"BC")
                  > and (AB"C") are concurrent in one point.

                  Note that if A',B',C' lie on the circumcircle, the circles A"B"C,
                  B"C"A, C"A"B have a common point.
                  I suspect that this common point lies on the circumcircle if and
                  only if the lines AA', BB', CC' concur.
                  Any idea?
                  Friendly. Jean-Pierre
                • Floor en Lyanne van Lamoen
                  Dear Jean-Pierre, [FvL] ... [JPE] ... If we have A and B , we can construct A , B , and then (A B C), and after that the common point P of (A B C) and the
                  Message 8 of 8 , Jan 7, 2005
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                    Dear Jean-Pierre,

                    [FvL]
                    > > To start the new year let me offer a little theorem:
                    > > If A'B'C' is a circumcevian triangle, and A"B"C" are the
                    > reflections of
                    > > A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
                    > (A"BC")
                    > > and (AB"C") are concurrent in one point.

                    [JPE]
                    > Note that if A',B',C' lie on the circumcircle, the circles A"B"C,
                    > B"C"A, C"A"B have a common point.
                    > I suspect that this common point lies on the circumcircle if and
                    > only if the lines AA', BB', CC' concur.
                    > Any idea?

                    If we have A' and B', we can construct A", B", and then (A"B"C), and after
                    that the common point P of (A"B"C) and the circumcircle. That means we have
                    the circles (AB"P)=(AB"C") and (A"BP)=(A"BC") as well. The point C" is thus
                    fixed as the second intersection of these circles (apart from P). So only
                    one point C", and consequently one point C' satisfies the condition. This
                    must be the circum-cevian trace of the intersection of AA' and BB'.

                    Kind regards,
                    Floor.
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